If the second term in the expansiona'$$13+aa-1nis$$ $$14a52$$ then the value of $$C3$$ $$nC2$$ n is

We have, $$T2=14a52$$ ⇒$$C1$$ $$na113n-1a321=14a52$$ ⇒$$nan-113+32=14a52$$ ⇒$$n=14$$ ⇒$$C3$$ $$nC2$$ $$n=C3$$ $$14C2$$ $$14=123=4$$

Binomial theorem for positive integral Index

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$If the second term in the expansion {(\sqrt[' 13]{a} + \frac{a}{\sqrt{a^{- 1}}})}^{n} is 14 a^{\frac{5}{2}} then the value of \frac{{}^{n}C_{3}}{{}^{n}C_{2}} is$

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Solution

$We have, T_{2} = 14 a^{\frac{5}{2}} \Rightarrow {}^{n}C_{1} {(a^{\frac{1}{13}})}^{n - 1} {(a^{\frac{3}{2}})}^{1} = 14 a^{\frac{5}{2}} \Rightarrow n a^{\frac{n - 1}{13} + \frac{3}{2}} = 14 a^{\frac{5}{2}} \Rightarrow n = 14 \Rightarrow \frac{{}^{n}C_{3}}{{}^{n}C_{2}} = \frac{{}^{14}C_{3}}{{}^{14}C_{2}} = \frac{12}{3} = 4$

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Binomial theorem for positive integral Index

Remember concepts with our Masterclasses.

If the second term in the expansiona'13+aa-1nis 14a52 then the value of C3 nC2 n is

We have, T2=14a52 ⇒C1 na113n-1a321=14a52 ⇒nan-113+32=14a52 ⇒n=14 ⇒C3 nC2 n=C3 14C2 14=123=4

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$If the second term in the expansion {(\sqrt[' 13]{a} + \frac{a}{\sqrt{a^{- 1}}})}^{n} is 14 a^{\frac{5}{2}} then the value of \frac{{}^{n}C_{3}}{{}^{n}C_{2}} is$